Math / Algebra

QuestionFor the functions $f(x)=\frac{x}{x+1}$ and $g(x)=\frac{11}{x}$ , find the composition $f \circ g$ and simplify your answer as much as possible. Write the dor notation. $(f \circ g)(x)=$ $\square$
Domain of $f \circ g$ : $\square$

Studdy Solution

STEP 1

What is this asking? We're taking two functions, $f(x)$ and $g(x)$ , and mashing them together!
We need to find the composition $f(g(x))$ , simplify it, and figure out what values of $x$ we're allowed to use. Watch out! Remember, $f(g(x))$ means we're putting $g(x)$ inside $f(x)$ .
It's like a function turducken!
Also, be careful with the domain; we need to consider the domains of both $f(x)$ and $g(x)$ .

STEP 2

1. Substitute $g(x)$ into $f(x)$
2. Simplify the composed function
3. Find the domain

STEP 3

Alright, let's start by remembering what our functions are: $f(x) = \frac{x}{x+1}$ and $g(x) = \frac{11}{x}.$ Now, to find $f(g(x))$ , we're going to take $g(x)$ and plug it right into $f(x)$ wherever we see an $x$ .
It's like substitution madness!

STEP 4

So, $f(g(x))$ becomes: $f(g(x)) = \frac{\frac{11}{x}}{\frac{11}{x} + 1}.$ See how we replaced the $x$ in $f(x)$ with the entire function $g(x)$ ?
Boom!

STEP 5

That fraction looks a little complicated, right?
Let's simplify it!
We can multiply the top and bottom of the big fraction by $x$ to get rid of those little fractions within the fraction.
Remember, multiplying by $\frac{x}{x}$ is the same as multiplying by one, so we're not changing the value of the expression.

STEP 6

So, we have: $f(g(x)) = \frac{\frac{11}{x} \cdot x}{(\frac{11}{x} + 1) \cdot x} = \frac{11}{11 + x}.$ Much cleaner, right?

STEP 7

Now, for the domain!
The domain of $f(g(x))$ is all the $x$ values we're allowed to use.
We need to consider two things: the domain of $g(x)$ and the domain of the simplified $f(g(x))$ .

STEP 8

For $g(x) = \frac{11}{x}$ , we can't have $x = 0$ because we can't divide by zero.

STEP 9

For our simplified $f(g(x)) = \frac{11}{11 + x}$ , we can't have $11 + x = 0$ , which means $x$ can't be $-11$ .

STEP 10

So, putting it all together, the domain of $f(g(x))$ is all real numbers except $x = 0$ and $x = -11$ .

STEP 11

$(f \circ g)(x) = \frac{11}{11 + x}$ Domain of $f \circ g$ : $x \ne 0$ and $x \ne -11$

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QuestionFor the functions $f(x)=\frac{x}{x+1}$ and $g(x)=\frac{11}{x}$ , find the composition $f \circ g$ and simplify your answer as much as possible. Write the dor notation. $(f \circ g)(x)=$ $\square$
Domain of $f \circ g$ : $\square$

Studdy Solution

STEP 1

STEP 2

1. Substitute $g(x)$ into $f(x)$
2. Simplify the composed function
3. Find the domain

STEP 3

Alright, let's start by remembering what our functions are: $f(x) = \frac{x}{x+1}$ and $g(x) = \frac{11}{x}.$ Now, to find $f(g(x))$ , we're going to take $g(x)$ and plug it right into $f(x)$ wherever we see an $x$ .
It's like substitution madness!

STEP 4

So, $f(g(x))$ becomes: $f(g(x)) = \frac{\frac{11}{x}}{\frac{11}{x} + 1}.$ See how we replaced the $x$ in $f(x)$ with the entire function $g(x)$ ?
Boom!

STEP 5

STEP 6

So, we have: $f(g(x)) = \frac{\frac{11}{x} \cdot x}{(\frac{11}{x} + 1) \cdot x} = \frac{11}{11 + x}.$ Much cleaner, right?

STEP 7

Now, for the domain!
The domain of $f(g(x))$ is all the $x$ values we're allowed to use.
We need to consider two things: the domain of $g(x)$ and the domain of the simplified $f(g(x))$ .

STEP 8

For $g(x) = \frac{11}{x}$ , we can't have $x = 0$ because we can't divide by zero.

STEP 9

For our simplified $f(g(x)) = \frac{11}{11 + x}$ , we can't have $11 + x = 0$ , which means $x$ can't be $-11$ .

STEP 10

So, putting it all together, the domain of $f(g(x))$ is all real numbers except $x = 0$ and $x = -11$ .

STEP 11

$(f \circ g)(x) = \frac{11}{11 + x}$ Domain of $f \circ g$ : $x \ne 0$ and $x \ne -11$

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Studdy Solution

STEP 1

STEP 2

1. Substitute g(x)g(x)g(x) into f(x)f(x)f(x)2. Simplify the composed function3. Find the domain

STEP 3

STEP 4

So, f(g(x))f(g(x))f(g(x)) becomes: f(g(x))=11x11x+1.f(g(x)) = \frac{\frac{11}{x}}{\frac{11}{x} + 1}.f(g(x))=x11​+1x11​​. See how we replaced the xxx in f(x)f(x)f(x) with the entire function g(x)g(x)g(x)?Boom!

STEP 5

STEP 6

So, we have: f(g(x))=11x⋅x(11x+1)⋅x=1111+x.f(g(x)) = \frac{\frac{11}{x} \cdot x}{(\frac{11}{x} + 1) \cdot x} = \frac{11}{11 + x}.f(g(x))=(x11​+1)⋅xx11​⋅x​=11+x11​. Much cleaner, right?

STEP 7

Now, for the **domain**!The domain of f(g(x))f(g(x))f(g(x)) is all the xxx values we're allowed to use.We need to consider two things: the domain of g(x)g(x)g(x) and the domain of the simplified f(g(x))f(g(x))f(g(x)).

STEP 8

For g(x)=11xg(x) = \frac{11}{x}g(x)=x11​, we can't have x=0x = 0x=0 because we can't divide by **zero**.

STEP 9

For our simplified f(g(x))=1111+xf(g(x)) = \frac{11}{11 + x}f(g(x))=11+x11​, we can't have 11+x=011 + x = 011+x=0, which means xxx can't be −11-11−11.

STEP 10

So, putting it all together, the domain of f(g(x))f(g(x))f(g(x)) is all real numbers *except* x=0x = 0x=0 and x=−11x = -11x=−11.

STEP 11

(f∘g)(x)=1111+x(f \circ g)(x) = \frac{11}{11 + x}(f∘g)(x)=11+x11​ Domain of f∘gf \circ gf∘g: x≠0x \ne 0x=0 and x≠−11x \ne -11x=−11

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1. Substitute $g(x)$ into $f(x)$
2. Simplify the composed function
3. Find the domain

So, $f(g(x))$ becomes: $f(g(x)) = \frac{\frac{11}{x}}{\frac{11}{x} + 1}.$ See how we replaced the $x$ in $f(x)$ with the entire function $g(x)$ ?
Boom!

So, we have: $f(g(x)) = \frac{\frac{11}{x} \cdot x}{(\frac{11}{x} + 1) \cdot x} = \frac{11}{11 + x}.$ Much cleaner, right?

Now, for the domain!
The domain of $f(g(x))$ is all the $x$ values we're allowed to use.
We need to consider two things: the domain of $g(x)$ and the domain of the simplified $f(g(x))$ .

For $g(x) = \frac{11}{x}$ , we can't have $x = 0$ because we can't divide by zero.

For our simplified $f(g(x)) = \frac{11}{11 + x}$ , we can't have $11 + x = 0$ , which means $x$ can't be $-11$ .

So, putting it all together, the domain of $f(g(x))$ is all real numbers except $x = 0$ and $x = -11$ .

$(f \circ g)(x) = \frac{11}{11 + x}$ Domain of $f \circ g$ : $x \ne 0$ and $x \ne -11$